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2 years ago

An agricultural manager requires

Engineering

1 answer:

sp2606 [1]2 years ago

3 0

C, being able to maintain legal information on grant programs

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What are the 3 dimensions that used in isometric sketches?

noname [10]

Answer:

The three dimensions shown in an isometric drawing are the height, H, the length, L, and the depth, D

Explanation:

An isometric drawing of an object in presents a pictorial projection of the object in which the three dimension, views of the object's height, length, and depth, are combined in one view such that the dimensions of the isometric projection drawing are accurate and can be measured (by proportion of scale) to draw the different views of the object or by scaling, for actual construction of the object.

5 0

3 years ago

An inventor claims to have developed a food freezer that at steady state requires a power input of 0.25 kW to extract energy by

Serga [27]

Answer:

investor claim is acceptable

Explanation:

given data

Win = 0.25 kW

Qc = 3000 J/s = 3kW

Th = 293 K

Tc = 270 K

solution

we get here coefficient of performance of cycle is

coefficient of performance = $\frac{Qc}{Win}$ ..................1

put here value and we get

coefficient of performance = $\frac{3}{0.25}$

coefficient of performance = 1.2

and

coefficient of performance of ideal refrigeration is

coefficient of performance = $\frac{Tc}{Th-Tc}$ ..................2

coefficient of performance = $\frac{270}{293-270}$

coefficient of performance = 11.74

and

we can see here that coefficient of performance of ideal refrigeration is is more than real cycle coefficient of performance

so investor claim is acceptable

6 0

3 years ago

When the electrical connection to the alternator from the battery is not tight, it can cause?

Oksi-84 [34.3K]

Answer:

affects the flow of electricity

Explanation:

A loose battery terminal affects the flow of electricity. There is less power going to the electrical systems and the vehicle will not start or start sluggishly. Also, a loose battery terminal causes the car's electrical components like navigation, car lights, and audio among others to dim or fail completely.

3 0

2 years ago

A ramjet operates by taking in air at the inlet, providing fuel for combustion, and exhausting the hot air through the exit. Th

UNO [17]

Answer:

15300 N

Explanation:

$\rho_i$ = Density of air at inlet

$\dfrac{m}{t}$ = Mass flow rate = 60 kg/s

$v_i$ = Inlet velocity = 225 m/s

$\rho_o$ = Density of gas at outlet = $0.25\ \text{kg/m}^3$

$A_i$ = Inlet area

$A_o$ = Outlet area = $0.5\ \text{m}^2$

Since mass flow rate is the same in the inlet and outlet we have

$\rho_iv_iA_i=\rho_ov_oA_o\\\Rightarrow v_o=\dfrac{\dfrac{m}{t}}{\rho_oA_o}\\\Rightarrow v_o=\dfrac{60}{0.25\times 0.5}\\\Rightarrow v_o=480\ \text{m/s}$

Thrust is given by

$F=\dfrac{m}{t}(v_o-v_i)\\\Rightarrow F=60\times (480-225)\\\Rightarrow F=15300\ \text{N}$

The thrust generated is 15300 N.

8 0

3 years ago

ASAE 1060 Steel wire (1 mm diameter) is coated with copper to form a composite with a diameter of 2mm. Use the following propert

k0ka [10]

Answer:

a) $E_{m}$ = 133.75 Gpa

b) Fnet = 560 N

c) thermal expansion of the composite material = 14.31 $10^{-6 }$ / °C

Explanation:

Solution:

a) Elastic Modulus of the composite:

Area of steel wire = $\frac{\pi }{4}$ x ( $0.001^{2}$ ) = 0.8 x $10^{-6}$ $m^{2}$

Area of Copper wire = $\frac{\pi }{4}$ x ( $0.002^{2}$ ) - 0.8 x $10^{-6}$ $m^{2}$

Area of Copper wire = 2.4 x $10^{-6}$ $m^{2}$

Young's Modulus of Composite mixture:

$E_{m}$ = $F_{st}$ $E_{st}$ + $F_{Cu}$ $E_{Cu}$ Equation 1

here,

$F_{st}$ = Stress in Steel

$F_{Cu}$ = Stress in Copper.

We know that,

F = P/A

F is inversely proportional to Area, so if area is large, stress will less and vice versa. So, Take

Ratio for area of steel = $\frac{0.8. 10^{-6} }{(0.8 + 2.4) .10^{-6} }$

Ratio for area of steel = $\frac{0.8}{3.2 }$ = 0.25

Similarly, for Copper,

Ratio for area of copper = $\frac{2.4. 10^{-6} }{(0.8 + 2.4) .10^{-6} }$

Ratio for area of copper = $\frac{2.4 }{3.2}$ = 0.75

Put these values in equation 1:

$E_{m}$ = $F_{st}$ $E_{st}$ + $F_{Cu}$ $E_{Cu}$

$E_{m}$ = (0.25) $E_{st}$ + (0.75) $E_{Cu}$

We are given that,

$E_{st}$ = 205 Gpa

$E_{Cu}$ = 110 Gpa

So,

$E_{m}$ = (0.25) (205 Gpa) + (0.75) (110 GPa)

$E_{m}$ = 51.25GPa + 82.5 Gpa

Hence, the Elastic Modulus of the composite will be:

$E_{m}$ = 133.75 Gpa

b) maximum force:

Fnet = Fst + Fcu

We know that F = (Yield Stress x Area)

F = fst x Ast + fcu x Acu

And we are given that,

Yield stress of Steel = 280 Mpa

Yield stress of Copper = 140 Mpa

And,

Ast = 0.8 x $10^{-6}$ $m^{2}$

Acu = 2.4 x $10^{-6}$ $m^{2}$

Just plugging in the values, we get:

F = (280 Mpa) (0.8 x $10^{-6}$ $m^{2}$ ) + (140 Mpa) (2.4 x $10^{-6}$ $m^{2}$ )

F = 224 + 336

Fnet = 560 N ( because Mpa = $10^{6}$ N/ $m^{2}$ )

So, it means the composite will carry the maximum force of 560N

c) Coefficient of Thermal Expansion:

Strain on both material is same upon loading so,

(ΔL/L)st = (ΔL/L)cu

by thermal expansion equation:

( $\alpha .$ ΔT + $\frac{F}{A}$ $. \frac{1}{Est}$ ) = $\alpha .$ ΔT + $\frac{F}{A}$ $. \frac{1}{Ecu}$ )

Where $\alpha$ = Coefficient of Thermal expansion

Here, fst = -fcu = F

and ΔT = 1°

So,

Plugging in the values, we get.

( 10 x $10^{-6}$ x (1) + $\frac{F}{0.8.10^{-6} } . \frac{1}{205 . 10^{9} }$ ) = ( 17 x $10^{-6}$ x (1) + $\frac{-F}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }$ )

Solving for F, we get:

F = 0.71 N

Here,

fst = F = 0.71 N (Tension on Heating)

fcu = -F = 0.71 N ( Compression on Heating )

So, the combined thermal expansion of the composite material will be:

(ΔL/L)cu = ( 17 x $10^{-6}$ x (1°) + $\frac{-0.71}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }$ )

(ΔL/L)cu = ( 17 x $10^{-6}$ x (1°) - 2.69 x $10^{-6}$

combined thermal expansion of the composite material = 14.31 $10^{-6 }$ / °C

4 0

3 years ago